Exercise on conformal metrics I am trying to prove that if we have 2 Levi-Civita Connections, $\nabla $ associated with the metric $\langle\cdot,\cdot\rangle$ and $\nabla^2$ associated with the metric $\langle\langle\cdot,\cdot\rangle\rangle$ , where
$$\langle\langle\cdot,\cdot\rangle\rangle = e^{2\rho}\langle\cdot,\cdot\rangle$$
where $\rho$ is a differentiable function, then 
$$\nabla²_XY= \nabla _XY+d\rho(X)Y+d\rho(Y)X - \langle X,Y\rangle \text{grad} \rho$$
where $\langle \text{grad} \rho,X\rangle =d\rho(X)$.
So I was thinking that the way of doing this would be to use the uniqueness of the Levi-Civita Connection and prove that the right side of the expression satisfies the 2 properties of the Levi-Civita Connection and we would be done.
I was able to do this for the symmetry of the connection but the other one I'm having some trouble with it, can anyone help me out? And is this the right way to think about this problem or is there a better way ? Thanks in advance.
 A: One way you can do this problem is with the Koszul Formula. Choose an arbitrary vector field $Z$, and use that formula to get
\begin{align}
2 \langle \nabla^2_X Y, Z \rangle_2 &= X \langle Y, Z \rangle_2 + Y \langle Z, X \rangle_2 - Z \langle X, Y \rangle_2\\
 &+ \langle [X, Y], Z \rangle_2 - \langle [Y, Z], X \rangle_2 - \langle [X, Z] , Y \rangle_2. \tag{*}\label{*}
\end{align}
Then
\begin{align}
X \langle Y, Z \rangle_2 = X(e^{2\rho}\langle Y, Z \rangle_1 ) &= e^{2\rho} X\langle Y, Z \rangle_1 + \langle Y, Z \rangle_1 X(e^{2\rho} )\\
 &= e^{2\rho} X\langle Y, Z \rangle_1 + \langle Y, Z \rangle_1 e^{2\rho} \cdot 2 d\rho(X).
\end{align}
We also have
$$\langle [X, Y], Z \rangle_2 = e^{2\rho} \langle [X, Y], Z \rangle_1.$$
Applying identities of this form to each term of the right-hand side in \eqref{*}, we get
\begin{align}
2 \langle \nabla^2_X Y, Z \rangle_2 = e^{2\rho} \cdot 2 \langle \nabla_X Y, Z   \rangle_1 + 2 d \rho(X)  e^{2\rho}\langle Y, Z \rangle_1 
+ 2 d \rho(Y)  e^{2\rho}\langle Z, X \rangle_1 - 2 e^{2\rho} \langle X, Y \rangle_1 \langle \text{grad} \rho, Z \rangle_1
\end{align}
But $\langle \nabla^2_X Y, Z \rangle_2 = e^{2\rho} \langle \nabla^2_X Y, Z \rangle_1$, so we can cancel $2e^{2\rho}$ from both sides and get
\begin{align}
\langle \nabla^2_X Y, Z \rangle_1 &= \langle \nabla_X Y, Z   \rangle_1 + d \rho(X)  \langle Y, Z \rangle_1 
+  d \rho(Y)  \langle Z, X \rangle_1 - \langle X, Y \rangle_1 \langle \text{grad} \rho, Z \rangle_1 \\
&= \langle \left(\nabla_X Y + d \rho(X) Y + d\rho(Y)X - \langle X, Y \rangle_1 \text{grad} \rho \right), Z \rangle_1. 
\end{align}
Since this last equation holds for all $Z$, the two vector fields are equal.
